Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Complex numbers

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img139.imageshack.us/img139/999/ksknl6.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

    Could someone confirm that the a) part is correct, and if it is, then what is the next step? :)
    And im not exactly sure what to do in the b) part.

    I would really appriciate if someone could help me out with this. :)

    Thanks in advance.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 21, 2008 #2

    HallsofIvy

    User Avatar
    Science Advisor

    NO, (a) is not correct because you haven't answered the question! I think you have misunderstood what is asked. If you were asked to find [itex]^4\sqrt{16}[/itex], you wouldn't answer "161/4"!

    Don't change from polar to rectangular form until after you have done the root! One important reason for using polar form for complex numbers is the fact that
    [tex][r (cos(\theta)+ i sin(\theta)]^n= r^n(cos(n\theta)+ i sin(n\theta))[/tex]

    For n= 1/4,
    [tex](7(cos(\pi/2)+ i sin(\pi/2))^{1/4}= 7^{1/4}(cos(\pi/8)+ i sin(\pi/8))[/tex]
    Also, since adding [itex]2\pi[/itex] to the argument doesn't change the complex number, and [itex]2\pi/4= \pi/2[/itex] another fourth root is
    [tex]7^{1/4}(cos(\pi/4+ \pi/2)+ i sin(\pi/4+ \pi/2))[/tex]
    yet another is
    [tex]7^{1/4}(cos(\pi/4+ \pi)+ i sin(\pi/4+ \pi))[/tex]
    and, finally,
    [tex]7^{1/4}(coS(\pi/4+ 3\pi/2)+ i sin(\pi/4+ 3\pi/2))[/tex]

    Those are the four fourth roots of [itex]7(cos(\pi/2)+ i sin(\pi/2))= 7i[/itex]
     
  4. Nov 21, 2008 #3
    Edit: Shows what I know. Listen to HallsofIvy instead
     
    Last edited by a moderator: Nov 21, 2008
  5. Nov 23, 2008 #4
    Last edited by a moderator: May 3, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook